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Artist 52 [7]
3 years ago
13

Student one used bowling ball A in a bowling game against Student 2, who used bowling ball B. Use Newton’s Three Laws of Motion

to explain the following:
Make a claim as to how Newton’s Three Laws of Motion can impact the bowling game for the 2 students. .

Use the image below and your background knowledge of how mass, acceleration, and force differ for the two bowling balls to support your claim, and provide reasoning as to how this evidence supports your claim.

Explain what variable the players could manipulate to maximize the force exerted on the pins.
Physics
1 answer:
Jlenok [28]3 years ago
8 0

Answer:

Newton’s Three Laws of Motion has a great impact.

Explanation:

Newton’s Three Laws of Motion has a great impact on the bowling game for the 2 students. When the student one throw ball to the student 2, the ball decrease its speed due to the gravity and opposing air. If these forces are removed from the system the ball will continue its motion till another force is applied on it. When the force applied to the ball it produces acceleration in the direction to the applied force. If the ball touches the ground it bounce back with equal force which is a reaction of the ground.

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Explanation:

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In our lab experiment on Ohm's law, the power supply is connected to a circuit containing one resistor, and a direct current of
lapo4ka [179]

Answer: (A) 3.0=A

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4 years ago
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4 years ago
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A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total
NeTakaya

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is 64.59^{\circ} South of East

(c) Average velocity magnitude is 570.88 km

(d) The direction of average velocity is 64.59^{\circ} South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, \vec{AB} = 446 km

Distance moved from B to C in South direction, \vec{BC} = - 939 km

Time taken to move from A to B, t = 43.0 min = 0.72 h

Time taken to move from B to C, t' = 1.10 h

Now,

(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:

AC = \sqrt{(AB)^{2} + (BC)^{2}}

AC = \sqrt{(446)^{2} + (- 939)^{2}} = 1039 km

(b) Direction of the displacement is given by:

tan\theta = \frac{\vec{BC}}{\vec{AB}}

\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}

64.59^{\circ} South of East

(c) Magnitude of the average speed is given by:

v_{avg} = \frac{AC}{t + t'}

v_{avg} = \frac{1039}{1.82} = 570.88 km/h

(d) The direction of the average velocity is the same as that of the displacement, i.e., 64.59^{\circ} South of East.

(e) The average speed of the [plane is given by:

v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}

v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h

6 0
3 years ago
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